Use the Laplace transform to solve the given differential equation subject to the indicated initial conditions. y''-7y'+6y =e^t + delta(t-3) + delta(t-5) y(0) = 0 y'(0) = 0

Step 1
Here we use table of Laplace transforms and some properties of Laplace transform.
Step 2
$y^{″}-7y^{\prime }+6y=e^t+\delta (t-3)+\delta (t-5)$
$y(0)=0$
$y^{\prime }(0)=0$
Apply laplace transform$\Rightarrow L\left\{y"\right\}-7L\left\{y^{\prime }\right\}+6L\left\{y\right\}=L\left\{e^t\right\}+L\{\delta (t-3)\}+L\{\delta (t-5)\}$
$\text{use }L\{\delta (t-c)\}=e^{-cs}$
$\Rightarrow (s^{2}Y(s)-sY(0)-y^{\prime }(0))-7(sY(s)-y(0))+6y(s)=\frac{1}{(s-1)}+e^{-3s}+e^{-5s}$
$\Rightarrow (s^2-7s+6)Y(s)=\frac{1}{(s-1)}+e^{-3s}+e^{-5s}$
$\Rightarrow Y(s)=\frac{1}{(s-1)(s^2-7s+6)}+\frac{e^{-3s}}{(s^2-7s+6)}+\frac{e^{-5s}}{(s^2-7s+6)}$
Now apply inverse laplace transform
$\Rightarrow L^{-1}\{Y(s)\}=L^{-1}\left\{\frac{1}{(s-1)(s^2-7s+6)}\right\}+L^{-1}\left\{\frac{e^{-3s}}{(s^2-7s+6)}\right\}+L^{-1}\left\{\frac{e^{-5s}}{(s^2-7s+6)}\right\}$
Now use paricl and use table of Laplace transforms
$\Rightarrow y(t)=-\frac{1}{6}t\cdot e^t-\frac{1}{6}{e}^{(}t-3)\cdot \delta (t-3)-\frac{1}{6}{e}^{t-5}\cdot \delta (t-5)$
here we use properties $L^{-1}\{F(s-c)\}=e^{ct}\cdot f(t)$
$L^{-1}\{e^{-cs}\cdot F(s)\}=y(t)\cdot f(t)$